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3 years ago

Please help! find the value of x in the isosceles triangle shown below. 4 6

Mathematics

1 answer:

nata0808 [166]3 years ago

6 0

Answer:

Step-by-step explanation:

Pythagorean theorem:

6^2 + (4/2)^2 = x^2

36 + 4 = x^2

40 = x^2

x = $\sqrt{40}$ = $2\sqrt{10}$

Answer is B

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inez waters her plants every two days she trim them every 15 days she did both today when will she do both again

lord [1]

She will do both again in 30 days.

She waters every two days and 2 is a even number.

She trims her plants every 15 days and 15 is an odd number.

When you add two odd numbers you get an even 15 + 15 = 30

2 × 15 = 30

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4 years ago

Aaron has a digital scale. He puts a marshmallow on The digital scale and it reads 7.2 grams. How much would you except 10 marsh

madreJ [45]

Answer:

72 grams

Step-by-step explanation:

If one marshmallow weighs 7.2 all you have to do is take that times 10, which would give you 72 grams as the answer.

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4 years ago

List all the 3-digit numbers that fit these clues .the hundreds digit is greater than7 .the tens is 1 more than the hundreds dig

djverab [1.8K]

In this question, there are 3 conditions that need to be met
1. The number is 3-digit
2. Hundreds digit >7 (number that will fulfill it would be 8 or 9)
3. The tens is 1 more than hundred( since the hundred possibilities is 8 or 9, then 9 in hundreds can't be used since no number higher than 9)
4. The one digit <2( that mean 0, 1)

Using the list above, you can make your possible number:
890
891

3 0

3 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

G(x) = -2x - 5<br> What is g(-3)?

kipiarov [429]

G-3=12
Hope it helps

5 0

4 years ago

Read 2 more answers